Leetcode 3 Longest Substring without Repeating Characters

Longest Substring without repeating characters

Question

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. 
             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

 

 

Naïve solution: brute force

The idea is to check every substring of the string, and use a set to check if the substring contains duplicate.

class Solution
{
public:
    int lengthOfLongestSubstring(string s)
    {
        int i, j, k;
        int longest = 0;
        for (i = 0; i<s.length() - 1; i++)
            for (j = i + 1; j<s.length(); j++)
            {
                set<char> substring;
                for (k = i; k <= j; k++)
                {
                    if (substring.find(s[k]) != substring.end())
                    {
                        longest = max(longest, int(substring.size()));
                        break;
                    }
                    else
                        substring.insert(s[k]);
                }
                //这个小心！ e.g. "pwwkew"
                longest = max(longest, int(substring.size()));
            }
​
        return longest;
    }
};

Sliding Window

A window is represented by $​[x,y)$​, containing a substring.

The idea is to use a sliding window to store the characters in current window, if s[j] is not in the set, then slide j further. Otherwise we find the maximum size of the current substring starting from i. Repeat for a new i.

class Solution 
{
public:
    int lengthOfLongestSubstring(string s) 
    {
        int i=0, j=0;
        int longest = 0;
        unordered_map <char, int> characters;
​
        while( i<s.length() && j<s.length() )
        {
            if(characters.find(s[j])==characters.end() || characters[s[j]]<i || characters[s[j]]>=j)
            {
                //此时map种没有这个值或者位置在window之外
                characters[s[j]] = j;
                j++;
            }
            else
            {
                if((j-i)>longest)
                    longest = j-i;
                i = characters[s[j]]+1;
                j = i;
            }
        }
​
        if((j-i)>longest)
            longest = j-i;
​
        return longest;
    }
};